We have \(\dfrac{x^4-1}{2x-2}=\dfrac{\left(x^2-1\right)\left(x^2+1\right)}{2\left(x-1\right)}=\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{2\left(x-1\right)}\)\(=\dfrac{\left(x+1\right)\left(x^2+1\right)}{2}=\dfrac{x^3+x^2+x+1}{2}\)

ĐKXĐ: \(2x-2\ne0\Leftrightarrow x\ne1\)

\(\dfrac{x^4-1}{2x-2}=\dfrac{(x^2)^2-1}{2x-2}=\dfrac{\left(x^2-1\right)\left(x^2+1\right)}{2\left(x-1\right)}\)

\(=\dfrac{\left(x-1\right)\left(x+1\right)(x^2+1)}{2\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x^2+1\right)}{2}\)

Ta có: \(\left(1+2a\right)\left(1-2a\right)-a\left(a+2\right)\)

\(=1-4a^2-a^2-2a\)

\(=-5a^2-2a+1\)

Nhầm yêu cầu đề không bạn?

\(\left(1+2a\right)\left(1-2a\right)-a\left(a+2\right)\)

\(=1-\left(2a\right)^2-a\left(a+2\right)\)

\(=1-4a^2-a\left(a+2\right)\)

\(=1-4a^2-a^2-2a\)

\(=-5a^2-2a+1\)

Ta có: \(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\)  (đk: x khác 1 và -1)

\(\Leftrightarrow\dfrac{\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=2\)

\(\Leftrightarrow\dfrac{x^2-3x-4+x^2+3x-4}{x^2-1}=2\)

\(\Rightarrow2x^2-8=2x^2-2\)

\(\Leftrightarrow0=6\) (vô nghiệm)

Vậy PT đã cho vô nghiệm

Bài 2:

a) \(5\left(x-1\right)\le6\left(x+2\right)\)

\(\Leftrightarrow5x-5\le6x+12\)

\(\Rightarrow x\ge-17\)

b) \(\dfrac{2x-1}{2}-\dfrac{x+1}{6}\ge\dfrac{4x-5}{3}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)-\left(x+1\right)}{6}\ge\dfrac{2\left(4x-5\right)}{6}\)

\(\Rightarrow6x-3-x-1\ge8x-10\)

\(\Leftrightarrow3x\le6\)

\(\Rightarrow x\le2\)

...

\(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\)

ĐKXĐ: \(x\ne\pm1\)

\(\Leftrightarrow\dfrac{\left(x-4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow x^2-3x-4+x^2+3x-4=2\left(x^2-1\right)\)

\(\Leftrightarrow2x^2-8-2x^2+2=0\)

\(\Leftrightarrow-6=0\) (Vô lí)

Vậy phương trình trên vô nghiệm.

(4x² +  8xy    - 3xy² )(-\(\dfrac{3}{4}\)x²y)

= -3x²y  - 6x³y² +\(\dfrac{9}{4}\)x³y³

4x³(2x² - x + 5) 5x

= 20x⁴(2x² - x + 5)

= 40x⁶ - 20x⁵ +100x⁴

\(\left(x-1\right)^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-2x+1-9\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow x^2-2x+1-9x^2-2x-9=0\)

\(\Leftrightarrow-8x^2-4x-8=0\)

\(\Leftrightarrow-4\left(2x^2+x+2\right)=0\)

\(\Leftrightarrow2x^2+x+2=0\)

Xét \(\Delta=1^2-4\cdot2\cdot2=-15< 0\)

⇒ Phương trình vô nghiệm

Vậy \(S=\varnothing\).

(x-1)² - 9 (x-1)² = 0

(x-1)².(1-9) =0

x-1 = 0

x = 1 

Ta có: \(\left(x-1\right)^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left[\left(x-1\right)-3\left(x+1\right)\right]\left[x-1+3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)

...

I'll let you draw the figure.

a) AM is a median of the triangle ABC. Therefore, M is the midpoint of BC.

E is symmetric to A through M, so M is the midpoint of AE.

Consider the quadrilateral ABEC, it has M is the midpoint of both AE and BC. Thus, ABEC is a parallelogram.

b) Consider the triangle ADE, M is the midpoint of AE, H is the midpoint of AD. Therefore, HM is the average line of this triangle. This means \(HM//DE\) or \(DE//BC\), which means BCED is a trapezoid.

Triangle ABD has the height BH, which is also a median. Thus, ABD must be an isosceles triangle, which means the height BH is also a bisector, or \(\widehat{ABH}=\widehat{DBH}\) or \(\widehat{ABC}=\widehat{DBC}\)

On the other hand, ABEC is a parallelogram. So, \(AB//CE\) and this leads to \(\widehat{ABC}=\widehat{ECB}\) (2 staggered angles of 2 parallel lines)

From these, we have \(\widehat{DBC}=\widehat{ECB}\left(=\widehat{ABC}\right)\)

The trapezoid BCED (DE//BC) has \(\widehat{DBC}=\widehat{ECB}\). Therefore, BCED must be an isosceles trapezoid.